Problem 23 - Single Variable Calculus

Question

Find the derivative of the function. $$ g(x)=\frac{e^{2 x}}{1+e^{-x}} $$

Step-by-Step Solution

Answer

The derivative of the function \(g(x) = \frac{e^{2x}}{1+e^{-x}}\) is $$ g'(x) = \frac{2e^{2x} + e^{x}}{(1+e^{-x})^{2}} $$

Step 1: Identify the Numerator and Denominator

Before applying the quotient rule, identify the numerator and the denominator. In this case: \( Num (x) = e^{2x} \) \( Den (x) = 1 + e^{-x} \)

Step 2: Derivative of the Numerator

Differentiate the numerator with respect to x using the chain rule. The chain rule states that if you have a composition of functions, the derivative of the composite function is the product of the derivative of the outer function and the derivative of the inner function. For the numerator, the outer function is the exponential function, and the inner function is 2x. \( \frac{d(Num)}{dx} = \frac{d(e^{2x})}{dx} = e^{2x} \cdot \frac{d(2x)}{dx} = 2e^{2x} \)

Step 3: Derivative of the Denominator

Differentiate the denominator with respect to x using the chain rule: \( \frac{d(Den)}{dx} = \frac{d(1+e^{-x})}{dx}= 0 - e^{-x} \cdot \frac{d(-x)}{dx} = e^{-x} \)

Step 4: Apply the Quotient Rule

Now, apply the quotient rule to find the derivative of the given function. The quotient rule states that if the function is of the form \(\frac{f(x)}{g(x)}\), then its derivative is given by: \( \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \) Using the values calculated in steps 2 and 3, substitute f'(x), f(x), g'(x), and g(x). \( g'(x) = \frac{(2e^{2x})(1+e^{-x}) - (e^{2x})(e^{-x})}{(1+e^{-x})^{2}} \)

Step 5: Simplify the Expression

By simplifying the expression, we will find the derivative of the provided function. \( g'(x) = \frac{2e^{2x}(1+e^{-x}) - e^{x}}{(1+e^{-x})^{2}} \) \( g'(x) = \frac{2e^{2x}+2e^{x} - e^{x}}{(1+e^{-x})^{2}} \) \( g'(x) = \frac{2e^{2x}+e^{x}}{(1+e^{-x})^{2}} \) Therefore, the derivative of the function g(x) = \(\frac{e^{2x}}{1+e^{-x}}\) is $$ g'(x) = \frac{2e^{2x} + e^{x}}{(1+e^{-x})^{2}} $$

Key Concepts

Quotient Rule

When it comes to finding the derivative of a quotient, the quotient rule is your go-to tool. It’s a rule that helps you differentiate expressions where you have one function divided by another function. Imagine you have a function in the form \( \frac{u(x)}{v(x)} \). To find its derivative, the quotient rule gives us a formula:

\[\frac{d}{dx} \left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]
Here's how it breaks down:

• \(u(x)\) is the numerator function.
• \(v(x)\) is the denominator function.
• \(u'(x)\) and \(v'(x)\) are their respective derivatives.
• You multiply \(u'(x)\) by \(v(x)\) and \(u(x)\) by \(v'(x)\) and then subtract the second product from the first.
• Finally, divide the result by \((v(x))^2\).

The quotient rule is very handy for keeping track of the roles played by the top and bottom functions of the fraction.

Chain Rule

Tackling derivatives often requires navigating through composite functions. That’s where the chain rule steps in. The chain rule is crucial when you have a function nested inside another function, making it a composite function.

Here’s the chain rule formula in action: if you have a function \( h(x) = f(g(x)) \), then the derivative \( h'(x) \) is:
\[h'(x) = f'(g(x)) \cdot g'(x)\]
Essentially, you:

• Identify the outer function \(f\) and the inner function \(g\).
• Differentiate the outer function using the inner function as the variable.
• Multiply by the derivative of the inner function.

For example, when differentiating \(e^{2x}\), the outer function is \(e^u\) and the inner function is \(u = 2x\). Using the chain rule, the derivative becomes \(e^{2x} \cdot 2\), simplifying to \(2e^{2x}\). Understanding the chain rule is essential because it lets you break down complexities in functions to find derivatives swiftly.

Exponential Functions

Exponential functions are special mathematical functions where the variable appears in the exponent. A classic example is \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828. These functions have unique properties, especially in calculus, making them pivotal in differential equations and applied mathematics.

Key Properties of Exponential Functions:

• The derivative of \(e^x\) is remarkably simple: it is \(e^x\). This is because the rate of change of an exponential function is proportional to its value, a key characteristic.
• Exponential growth is represented by expressions of the form \(e^{kx}\), where \(k\) is a constant. The derivative in this case is \(ke^{kx}\), thanks to the chain rule.

When dealing with composite exponential functions like \(e^{2x}\) or \(e^{-x}\), you apply the chain rule to find the derivative. Remember that the exponential function maintains its form even during differentiation, making it a smooth and predictable process. Their predictability makes them invaluable in models involving population growth, radioactive decay, and more.